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Chapter 1: Problem 79

A company that produces tracking devices for computer disk drives finds that if it produces \(x\) devices per week, its costs will be \(C(x)=180 x+16,000\) and its revenue will be \(R(x)=-2 x^{2}+660 x\) (both in dollars). a. Find the company's break-even points. b. Find the number of devices that will maximize profit, and the maximum profit.

Short Answer

Expert verified
Break-even at 30 devices. Maximum profit is \$12,800 at 120 devices.

Step by step solution

01

Find Break-even Points

The break-even point occurs when cost equals revenue, so we solve the equation \( C(x) = R(x) \). Given:\[ C(x) = 180x + 16,000 \]\[ R(x) = -2x^2 + 660x \]Setting them equal to each other:\[ 180x + 16,000 = -2x^2 + 660x \]Rearrange:\[ 2x^2 + (660 - 180)x - 16,000 = 0 \]\[ 2x^2 + 480x - 16,000 = 0 \]This quadratic equation can be solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 2\), \(b = 480\), and \(c = -16,000\).
02

Calculate the Discriminant and Roots

First, calculate the discriminant:\[ b^2 - 4ac = 480^2 - 4 \times 2 \times (-16,000) \]\[ = 230,400 + 128,000 = 358,400 \]Since the discriminant is positive, there are two real roots. Calculate the roots:\[ x = \frac{-480 \pm \sqrt{358,400}}{4} \]\[ x = \frac{-480 \pm 598.66}{4} \]Calculate the values of \(x\):\[ x_1 = \frac{-480 + 598.66}{4} \approx 29.66 \]\[ x_2 = \frac{-480 - 598.66}{4} \approx -269.66 \]Since producing a negative number of devices is not feasible, the break-even point in terms of the number of devices is approximately 30 (rounding to the nearest whole number).
03

Define Profit Function

Profit \( P(x) \) is defined as revenue minus cost:\[ P(x) = R(x) - C(x) \]Substitute the given expressions for \(R(x)\) and \(C(x)\):\[ P(x) = (-2x^2 + 660x) - (180x + 16,000) \]Simplify:\[ P(x) = -2x^2 + 480x - 16,000 \]
04

Find the Maximum Profit

The profit function \(-2x^2 + 480x - 16,000\) is a downward-opening parabola. The maximum profit occurs at the vertex of the parabola.The vertex for a parabola \( ax^2 + bx + c \) occurs at \( x = \frac{-b}{2a} \).Substitute \(a = -2\), \(b = 480\):\[ x = \frac{-480}{2(-2)} = 120 \]The number of devices that maximize profit is 120.
05

Calculate the Maximum Profit

Substitute \(x = 120\) back into the profit function:\[ P(120) = -2(120)^2 + 480(120) - 16,000 \]Calculate each term:\[ = -2(14,400) + 57,600 - 16,000 \]\[ = -28,800 + 57,600 - 16,000 \]\[ = 12,800 - 16,000 \]\[ = 12,800 \]Therefore, the maximum profit is \$12,800.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Formula
The quadratic formula is a fundamental tool in algebra used for finding the roots of quadratic equations of the form \(ax^2 + bx + c = 0\). To solve such equations, we employ the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
  • Components: Here, \(a\), \(b\), and \(c\) are coefficients of the quadratic equation, while \(x\) are the solutions or roots.
  • Discriminant: The expression \(b^2 - 4ac\) is known as the discriminant. It determines the nature of the roots:
    • If the discriminant is positive, there are two distinct real roots.
    • If it is zero, there is exactly one real root (a repeated root).
    • If it is negative, the equation has two complex roots.
  • Application: In the context of this problem, the quadratic formula is used to find the break-even points where costs equal revenues. By substituting the specific values from the problem into the formula, we found real roots corresponding to feasible production levels.
This straightforward formula helps in breaking down complex problems into simple calculations, facilitating easier understanding and prompt resolution of quadratic equations.
Profit Maximization
Profit maximization is a critical concept in business and economics, which involves determining the level of output that brings the highest possible profit to a company. The path to maximizing profit involves both understanding and utilizing various mathematical functions.In our case, the profit function is given by:\[P(x) = R(x) - C(x)\]Where \(P(x)\) is profit, \(R(x)\) is revenue, and \(C(x)\) is cost.
  • Formulation: We formulated the profit function by subtracting the cost function from the revenue function. This yields:\[P(x) = -2x^2 + 480x - 16,000\]
  • Types of Functions: The profit function is represented by a downward-opening parabola, indicating that there is a maximum point on the curve where profit is maximized. This is typical of profit functions that are quadratic, showcasing a clear peak point at the vertex.
  • Calculation: Finding the maximum profit involves determining the vertex of the parabola, which is achieved by applying a specific formula for the vertex of a quadratic equation.
Profit maximization requires understanding such functions and employing calculus or algebra to determine the point where the highest profit is achieved.
Vertex of a Parabola
The vertex of a parabola in quadratic equations represents an essential point that can either minimize or maximize a given function. In a standard form quadratic equation \(ax^2 + bx + c\), the vertex gives us the point where the parabola changes direction.For a function like \(-2x^2 + 480x - 16,000\), deriving the vertex helps determine the maximum profit, since the parabola opens downwards.
  • Vertex Formula: The formula \(x = \frac{-b}{2a}\) is used to locate the x-coordinate of the vertex, where \(a\) and \(b\) are coefficients from the quadratic expression.
  • Maximum or Minimum: Depending on the orientation of the parabola:
    • If \(a > 0\), the parabola opens upwards, and the vertex is the minimum point.
    • If \(a < 0\), the parabola opens downwards, and the vertex is the maximum point, which was the scenario in our example.
  • Practical Application: From the problem, the vertex was found to occur at \(x = 120\). This implies that producing and selling 120 devices maximizes the profit, allowing the company to achieve its financial objectives more effectively.
The vertex acts as a critical guide for decision-making in profit strategies, showing exactly where resources should be focused for optimum results.

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